For ~k = 0~, just answer 0 if ~x_1 = x_2~ or ~y_1 = y_2~ and -1 otherwise.

For ~n, m \le 300~, ~q \le 10~, for each query we can simulate a path in each direction from ~(x_1, y_1)~ and see if it reaches ~(x_2, y_2)~ and count turns along the way. It might end up in an infinite loop; we can force it to stop after ~10^6~ moves, because it's impossible for the distance to ~(x_2, y_2)~ to be that high.

For ~n, m \le 300~, we can imagine a graph where each pair of ~(x, y)~ and a direction is considered a vertex, and there is an edge from one vertex to another if, after going from ~(x, y)~ in the direction given by the initial vertex, our next move will be going from ~(x, y)~ of the next vertex in the direction given by the next vertex. Notice that each vertex has outdegree ~1~ and outdegree ~1~, so we have a graph which is a union of cycles. Now we can easily see if ~2~ vertices are reachable from one another by checking if they're in the same cycle, and also find the number of turns between them. We can do that for all ~4~ vertices that go out of ~(x_1, y_1)~ and all ~4~ vertices that go into ~(x_2, y_2)~ and see which pair gives the smallest number of turns.

For the full solution, notice that, outside of the case where ~x_1 = x_2~ and row ~x_1~ doesn't have anything in it or ~y_1 = y_2~ and column ~y_1~ doesn't have anything in it, the only ~(x, y)~ that you need to make vertices for are the ~k~ given ~(x, y)~. With that, finding the minimum number of turns becomes quite easy, as every vertex in the cycle represents one turn.